Question

If $f\left(x\right)=\frac{4^x}{4^x+2}$, where $x\in Q$ then $f\left(\frac{1}{2007}\right)+f\left(\frac{2}{2007}\right)+....+f\left(\frac{2006}{2007}\right)$ equal to

• A. $1003$
• B. $2006$
• C. $2007$
• D. $noneofthese$

Answer 1

Answer: (A)

$1003$

$\begin{array}{c}\text{Solution:}f\left(x\right)=\frac{4^x}{4^x+2},x\in Q\\ f\left(\frac{1}{2001}\right)+f\left(\frac{2}{2001}\right)+\dots +f\left(\frac{2006}{2007}\right)=?\end{array}$

$\begin{array}{cc}\Rightarrow f(1-x)=\frac{4^{1-x}}{4^{1-x}+\sqrt{4}}& =\frac{\frac{4}{4^x}}{\frac{4}{4^x}+\sqrt{4}}=\frac{4}{4+4^x\sqrt{4}}\\ f\left(x\right)+f(1-x)& =\frac{4^x}{4^x+\sqrt{4}}+\frac{4}{\sqrt{4}(\sqrt{4}+4^x)}\\ & =\frac{\sqrt{4}\left(4^x\right)+4}{\sqrt{4}(\sqrt{4}+4^x)}=1\end{array}$ $\begin{array}{cc}& f\left(\frac{1}{2007}\right)+f\left(\frac{2}{2007}\right)+\dots +f\left(\frac{2006}{2007}\right)\\ \Rightarrow & f\left(\frac{1}{2001}\right)+f\left(\frac{2006}{2007}\right)+f\left(\frac{2}{2007}\right)+f\left(\frac{2005}{2007}\right)\dots +\left(\frac{1003}{2007}\right)f\left(\frac{1004}{2007}\right)\\ \Rightarrow & f\left(\frac{1}{2007}\right)+f(1-\frac{1}{2007})+\dots .+\frac{1003}{2007})+f(1-\frac{1003}{2007})\\ \Rightarrow & 1003\end{array}$

Hence. (A) is the correct option.